Question

calculator may be used to determine the final numeric value, but show all steps in solving without a calculator up to the final calculation. the surface area a and volume v of a spherical balloon are related by the equationA³ - 36πV² where A is in square inches and Vis in cubic inches. If a balloon is being inflated with gas at the rate of 18 cubic inches per second, find the rate at which the surface area of the balloon is increasing at the instant the area is 153.24 square inches and the volume is 178.37 cubic inches.

Answer 1

10.309 in²/s

Explanation:

Given A³ = 36πV² and V' = 18 in³/s, you want to know A' when A=153.24 in² and V=178.37 in³.

Differentiation

Using implicit differentiation, we have ...

3A²·A' = 36π·2V·V'

A' = (36π·2)/3·V/A²·V' = 24πV/A²·V'

A' = 24π·(178.37 in²/(153.24 in²)²·18 in³/s

A' ≈ 10.309 in²/s

The surface area is increasing at about 10.309 square inches per second.

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Additional comment

There are at least a couple of ways a calculator can be used to find the rate of change. The first attachment shows evaluation of the expression we derived above. The second attachment shows the rate of change when the area is expressed as a function of the volume.

The result rounded to 5 significant figures is the same for both approaches.